Given, $\hat{i}+4\hat{j}+3\hat{k},\hat{i}+2\hat{j}+3\hat{k},3\hat{i}+2\hat{j}+\hat{k}$ are position vectors of $A,B,C$ respectively.

$D$ is mid point of $BC$, then position vector of $D$ is $\frac{\left(1+3\right)\hat{i}+\left(2+2\right)\hat{j}+\left(3+1\right)\hat{k}}{2}=2\hat{i}+2\hat{j}+2\hat{k}$

Similarly, position vector of $E$ is $\frac{\left(1+3\right)\hat{i}+\left(4+2\right)\hat{j}+\left(3+1\right)\hat{k}}{2}=2\hat{i}+3\hat{j}+2\hat{k}$

Then, $\overrightarrow{DE}=\left(2\hat{i}+3\hat{j}+2\hat{k}\right)-\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=\hat{j}$.